Question

Assume that 4 bits have been borrowed. Identify the subnetwork addresses: (choose 3) a. 192.168.14.8 b. 192.168.14.16 c. 192.168.14.24 d. 192.168.14.32 e. 192.168.14.148 f. 192.168.14.208

Answer 1

We know this is a Class C address, so we've got eight host bits to begin with: 00000000

The question says four bits have been borrowed for subnetting, so we'll put the subnet bits in bold : 00000000

With all subnet bits set to zero, we get this: 192.168.14.0. That's the first network number, but it's not in the list.

Now set the smallest subnet bit to one - 00010000 - we get 192.168.14.16. That's one on the list.

The next smallest subnet bit combination is 00100000 - we get 192.168.14.32. That's two on the list.

The next smallest subnet bit combination is 00110000 - we get 192.168.14.48. You probably notice a pattern here. :) Each subsequent network number is going up by 16 in that last octet. Going up by 16 from here, we get... 48, 64, 80, 96, 112, 128, 144, 160, 176, 192, 208 --- 208 is the final choice from that list.

Therefore the answers are:

b)192.168.14.16

d)192.168.14.32

f) 192.168.14.208

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